(x²-3x+1)(x²+3x+2)( x²-9x+20)
=(x²-3x+1)(x+1)(x+2)(x-4)(x-5)
=(x²-3x+1)(x+1)(x-4)(x+2)(x-5)
==(x²-3x+1)(x²-3x-4)(x²-3x-10)=-30
令x²-3x-4=t
则有(t+5)(t-6)t=-30
t³-t²-30t+30=0
即(t-1)(t²-30)=0
t=1或t=±√30
当t=1,x²-3x-5=0 x=(3±√29)/2
当t=√30,x²-3x-5-√30=0 x=[3±√(29+4√30)]/2=[3±√(√24+√5)²]/2=[3±(2√6+√5)]/2
当t=-√30,x²-3x-5+√30=0,x=[3±√(29-4√30)]/2=[3±(2√6-√5)]/2
综上
x=(3±√29)/2或[3±(2√6±√5)]/2